৪৯তম বিসিএস ⎯ পরিসংখ্যান [৯৮১]

CLT holds regardless of symmetry; only finite variance is required.

The Chi-square test of independence is used with contingency tables to check whether two categorical variables are independent or associated.

Correlation coefficient, on the other hand, is used for continuous bivariate data.

Source: Business Statistics- Mk Roy

A contingency table is used to display the frequency distribution of two categorical variables.

For example, a table showing the relationship between gender (male/female) and smoking habit (yes/no).

Source: Live MCQ class lecture (Probability basic previous lecture)

Source: Business Statistics (SP Gupta, MP Gupta) 

Bivariate data involves the analysis of two variables simultaneously to study the relationship between them.

For example, height and weight of students.

This is different from univariate (one variable) or multivariate (more than two variables).

Number of females who like tea = 30

Total number of people = 100

So, Probability = 30/100=0.30

Source: Live MCQ probability basic class lecture

The expected value is the theoretical mean of a random variable.

It does not always equal the most frequent (mode) or a possible outcome, but it represents the average value one would expect in the long run if the experiment were repeated many times.

Variance is a squared measure of spread, so it cannot be negative.

E(2X+3Y) = 2E(X)+3E(Y)

= 2(10)+3(15) = 20+45 = 65

E(XY)

= E(X)E(Y)

only if X and Y are independent.

By linearity of expectation:

E(2X+3)
= 2E(X)+3
= 2(4)+3
= 11

A fair coin has equal probability 0.5 for head (1) and tail (0).

CLT justifies widespread use of normal distribution in inference.

The Central Limit Theorem (CLT) is important because it explains why the normal distribution is so widely used in statistics: regardless of the population’s original distribution, the sampling distribution of the sample mean approaches a normal distribution as the sample size becomes large.

This makes it possible to apply normal-based methods (confidence intervals, hypothesis tests, probability approximations) in many real-world problems where the population distribution is unknown

Source: Live MCQ lecture Slide

CLT → sample mean of large samples ≈ normal, regardless of parent distribution.

Geometric is discrete distribution with memoryless property.

Only Geometric (discrete) and Exponential (continuous) have memoryless property.

Mean and variance of Poisson are both λ.

Central Limit Theorem → sums/averages of many variables → normal.

Poisson models count of rare events in a fixed interval.

According to characteristics of CDF,
CDF → non-decreasing, bounded between 0 and 1, tends to 1 at infinity.

Source: Live MCQ lecture slide

Bernoulli Variance = pq = 0.7(0.3) = 0.21.

In normal distribution, ~95% of values fall within 2 standard deviations.

Can’t have more successes in the sample than sample size. Max = 5.

Discrete RV = countable values (0,1,2,...).

Continuous RV = infinite, uncountable values in an interval.

According to definition, Var(X) = E[X²]−(E[X])². So, Variance is not equal to square of expectation.

For a fair die, possible outcomes = 1, 2, 3, 4, 5, 6 each with probability 1/6​.

E(X)= (1+2+3+4+5+6)/6=21/6=3.5S

So, the expected value = 3.5.

The formula for expected value:

E(X) = ∑[x⋅P(X)]

= (0)(0.1)+(1)(0.3)+(2)(0.4)+(3)(0.2) 

= 0+0.3+0.8+0.6=1.7

Hypergeometric = sampling without replacement.

option a-> Binomial distribution

Option c-> poisson distribution

Option d-> Geometric distrbution

E(X) can be negative depending on Random Variable.

CDF is monotonic non-decreasing.

With large n and not too small p, binomial ≈ normal.

When n large, p small → Poisson approx.

Geometric Distribution:

Binomial → independent with replacement,

Hypergeometric → without replacement.

A random variable maps outcomes (like heads, tails) to numbers (0,1).

It is not always between 0 and 1 (e.g., dice can take values 1–6).

Var(aX+b)=a².Var(X)

Here, a=2,Var(X)=9

Var(2X+3) = 2².9 = 36

Explanation: By definition, P(a≤X≤b) = F(b)−F(a)

Explanation: According to characteristics PDF cannot be negative.

Explanation: CDF either increases or stays flat but never decreases.

*IF INTERGATION OF A PDF GIVES 1, THEN THAT IS A VALID PDF.
Source: Business Statistics, Md. Abdul Aziz.
 Business Statistics, Roy.

Source: Business Statistics- Md Abdul Aziz

For uniform Distribution,

Var(X) = (b−a)²/12

​Here a=0, b=1 ⇒1²/12=1/12

Var(X) = np(1−p) = 10×0.5×⋅0.5 = 2.5

For Exponential distribution,

E(X)=1/λ

 Expectation is linear:

E(aX+b) = aE(X)+b = aμ+b

For Poisson distribution,

E(X) = λ = 3, Var(X) = λ = 3.

So, answer is 0.

For Bernoulli distribution:

E(X) = p, Var(X) = p(1−p).